Question: The 2D Infinite Well: In two dimensions the Schrödinger equation is

$\mathbf{(}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{y}}^{\mathbf{2}}}\mathbf{)}\mathit{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{(}\mathbf{E}\mathbf{-}\mathbf{U}\mathbf{)}}{{\mathbf{h}}^{\mathbf{2}}}\mathit{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$

(a) Given that U is a constant, separate variables by trying a solution of the form $\mathit{\psi }\left(x,y\right)\mathbf{=}\mathit{f}\left(x\right)\mathit{g}\left(y\right)$, then dividing by$\mathit{f}\left(x\right)\mathit{g}\left(y\right)$ . Call the separation constants C_X and C_Y .

(b) For an infinite well

role="math" localid="1659942086972" $$\begin{gathered} \mathit{U}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{y}\mathbf{<}\mathbf{L} \\ \mathbf{\infty }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{r}\mathbf{w}\mathbf{i}\mathbf{s}\mathbf{e} \end{gathered}$$

What should f(x) and g(y) be outside the well? What functions should be acceptable standing wave solutions f(x) for g(y) and inside the well? Are C_X and C_Y positive, negative or zero? Imposing appropriate conditions find the allowed values of C_X and C_Y .

(c) How many independent quantum numbers are there?

(d) Find the allowed energies E .

(e)Are there energies for which there is not a unique corresponding wave function?

The Schrodinger equation in two dimensionsis

$\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)\psi \left(x,y\right)=\frac{-2m\left(E-U\right)}{h^2}\psi \left(x,y\right)······\left(1\right)$$\left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)\psi \left(x,y\right)=\frac{-2m\left(E-U\right)}{h^2}\psi \left(x,y\right)······\left(1\right)$

The potential of a 2D infinite well is

$$\begin{gathered} U=\left\{00<x<L,0<y<L \\ \infty \text{otherwise} \\ \right. \end{gathered}$$

The general solution to the differential equation

$\frac{d^{2}X}{d{x}^2}=-k^{2}X·······\left(2\right)$

is of the form

$X=A\sin \left(kx\right)+B\cos \left(kx\right)······\left(3\right)$ $X=A\sin \left(kx\right)+B\cos \left(kx\right)······\left(3\right)$

where A and B are constants

Let

$\psi \left(x,y\right)=f\left(x\right)g\left(y\right)$

Substitute this in equation (1) to get

$$\begin{gathered} \left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \\ g\left(y\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x^2}+f\left(x\right)\frac{{\partial }^{2}g\left(y\right)}{\partial y^2}=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \end{gathered}$$$$\begin{gathered} \left(\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}\right)f\left(x\right)g\left(y\right)=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \\ g\left(y\right)\frac{{\partial }^{2}f\left(x\right)}{\partial x^2}+f\left(x\right)\frac{{\partial }^{2}g\left(y\right)}{\partial y^2}=\frac{-2m\left(E-U\right)}{h^2}f\left(x\right)g\left(y\right) \end{gathered}$$

Divide throughout by f(x) g(y) to get

$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial x^2}+\frac{1}{g\left(y\right)}\frac{{\partial }^{2}g\left(y\right)}{\partial y^2}=\frac{-2m\left(E-U\right)}{h^2}$$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial x^2}=-C_x······\left(4\right)$

Call

$\frac{1}{f\left(x\right)}\frac{{\partial }^{2}f\left(x\right)}{\partial x^2}=-C_x······\left(4\right)$

and

$\frac{1}{g\left(y\right)}\frac{{\partial }^{2}g\left(y\right)}{\partial y^2}=-C_y······\left(5\right)$

Finally, $C_x+C_y=\frac{2m\left(E-U\right)}{{\hslash }^2}······\left(6\right)$

Hence, the solution is $C_x+C_y=\frac{2m\left(E-U\right)}{h^2}$

Since the well is infinite there should be no probability of any particle of staying outside it. Hence f(x) and g(y) should be 0 outside the wall.

The equations (IV) and (V) are of the form (II). Their general solutions inside the well are of the form of equation (III) as follows

$$\begin{gathered} f\left(x\right)=A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right) \\ g\left(y\right)=C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right) \end{gathered}$$

Here A and B are functions of x and C and D are functions of y. In general C_x and C_Y can be positive, negative or zero.

The complete wave function becomes

$\psi \left(x,y\right)=\left\{A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right)\right\}\left\{C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right)\right\}$

The first set of boundary conditions are

$$\begin{gathered} \psi \left(x,0\right)=0 \\ \psi \left(0,y\right)=0 \end{gathered}$$

Substitute these to get

$$\begin{gathered} \psi \left(x,0\right)=\left\{A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right)\right\}D \\ =0 \\ D=0 \end{gathered}$$

and

$$\begin{gathered} \psi \left(0,y\right)=B\left\{C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right)\right\} \\ =0 \\ B=0 \end{gathered}$$

Combine C to A and get

$\psi \left(x,y\right)=A\sin \left(\sqrt{C_x}x\right)\sin \left(\sqrt{C_y}y\right)$

The second set of boundary conditions are

$$\begin{gathered} \psi \left(x,L\right)=0 \\ \psi \left(L,y\right)=0 \end{gathered}$$

The first one gives

$$\begin{gathered} A\sin \left(\sqrt{C_x}x\right)\sin \left(\sqrt{C_y}L\right)=0 \\ \sin \left(\sqrt{C_y}L\right)=0 \\ \sqrt{C_y}L=n_y\pi n_y=1,2.... \\ \sqrt{C_y}=\frac{n_y\pi }{L} \\ {C}_y={\left(\frac{n_y\pi }{L}\right)}^2 \end{gathered}$$

The second one gives

$$\begin{gathered} A\sin \left(\sqrt{C_x}L\right)\sin \left(\sqrt{C_y}y\right)=0 \\ \sin \left(\sqrt{C_x}L\right)=0 \\ \sqrt{C_x}L=n_x\pi n_x=1,2.... \\ \sqrt{C_x}=\frac{n_x\pi }{L} \\ {C}_x={\left(\frac{n_x\pi }{L}\right)}^2 \end{gathered}$$

The final wave function becomes

$\psi \left(x,y\right)=A\sin \left(\frac{n_x\pi x}{L}\right)\sin \left(\frac{n_y\pi y}{L}\right)$

Hence, The functions andare zero outside the wall. Inside the wall the general solutions are

$$\begin{gathered} f\left(x\right)=A\sin \left(\sqrt{C_x}x\right)+B\cos \left(\sqrt{C_x}x\right) \\ g\left(y\right)=C\sin \left(\sqrt{C_y}y\right)+D\cos \left(\sqrt{C_y}y\right) \end{gathered}$$

Allowed values of the constants are

$$\begin{gathered} C_x={\left(\frac{n_x\pi }{L}\right)}^2{n}_x=1,2,3,.. \\ {C}_y={\left(\frac{n_y\pi }{L}\right)}^2{n}_y=1,2,3,.. \end{gathered}$$

From equation (VI) and the value of potential inside the well

$$\begin{gathered} C_x+C_y=\frac{2mE}{h^2} \\ \frac{n_x^2{\pi }^2}{L^2}+\frac{n_y^2{\pi }^2}{L^2}=\frac{2mE}{h^2} \\ E=\frac{h^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_y^2\right) \end{gathered}$$

Thus the allowed values of energy are $\frac{h^2{\pi }^2}{2m{L}^2}\left(n_x^2+n_y^2\right)$ with two independent quantum numbers n_x and n_y .

The energies for which n_x and n_y are not equal have no unique corresponding wave functions.

Separate set of quantum numbers that have the same $n_x^2+n_y^2$ have the same energy. For example the wave functions $\psi \left(x,y\right)=A\sin \left(\frac{\pi x}{L}\right)\sin \left(\frac{2\pi y}{L}\right)$ and $\psi \left(x,y\right)=A\sin \left(\frac{2\pi x}{L}\right)\sin \left(\frac{\pi y}{L}\right)$ have quantum numbers (1 , 2) and (2 ,1) but they both have energy $\frac{{\hslash }^2{\pi }^2}{2m{L}^2}\left(1^2+2^2\right)=\frac{5{\hslash }^2{\pi }^2}{2m{L}^2}$.